Provided that $\vec{r}_1$ and $\vec{r}_2$ are a quarter turn away from each other, a circle can be described using the equation $\vec{r}(t)=\left(\vec{r}_1-\vec{r}_c\right)\cos(t)+\left(\vec{r}_2-\vec{r}_c\right)\sin(t)$, which involves only elementary functions as $\sin(t)$ and $\cos(t)$ are both elementary functions that can be expressed using exponential functions with imaginary numbers. Using the above equation for a circle $\vec{r}'(t)$ and $\vec{r}''(t)$ are perpendicular to each other and $\|\vec{r}'(t)\|$ is a constant.
I know there are other curves, for which $\vec{r}'(t)$ and $\vec{r}''(t)$ are perpendicular and $\|\vec{r}'(t)\|$ is constant, that can be expressed using differential equations, and such curves require $\vec{r}(t)$ to have at least 2 components.
Are there are any other curves for which $\vec{r}'(t)$ and $\vec{r}''(t)$ are perpendicular and $\|\vec{r}'(t)\|$ is constant that can be expressed entirely with elementary functions?
Assume euclidean geometry, and cartesian coordinates.
This might not be a perfectly complete answer, but I think it’ll help get you on the right track.
Any sufficiently differentiable curve r that satisfies $|\mathbf{r’}(t)| = c$ will also satisfy your other condition. To see why, notice that $| \mathbf{r’}(t)|^2 = \mathbf{r’}(t) \cdot \mathbf{r’}(t) = c^2.$ Taking a derivative on both sides, by the product rule we have that $$2\mathbf{r’}(t) \cdot \mathbf{r’’}(t) = 0,$$ so r’ and r’’ are orthogonal.
So to try to answer your question, there are other curves where the derivative has constant magnitude, but it’s up to you whether the first and second derivatives count as “perpendicular”. For example, any constant curve $\mathbf{r}(t) = c$ has zero first derivative everywhere and zero second derivative everywhere, so the first and second derivatives satisfy $\mathbf{r’}(t) \cdot \mathbf{r’’}(t) = 0.$ In the same sense, any line in space will have constant first derivative and zero second derivative. If you consider curves where the first derivative itself is constant, then the answer is yes, there are curves other than circles, expressed with elementary functions, whose first derivative has constant magnitude and satisfy $\mathbf{r’}(t) \cdot \mathbf{r’’}(t) = 0$.
If you want $\mathbf{r’}(t)$ to not be constant but to still have constant magnitude, then you’d need to find a set of functions $f_i(t)$ such that $\sum f_i^2 = c^2$. Then $\mathbf{r}(t) = \langle \int f_1(t) \, dt, \int f_2(t)\, dt, \dots, \int f_n(t) \, dt \rangle$ is a curve with $n$ components that satisfies your conditions. The only elementary functions $f_i$ that easily come to my mind are constant functions (and the sine and cosine when $n = 2$).